Question

Which of the following is not favourable for $S{O}_3$ formation
$2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right);△H=-45.0kcal$

• A. High pressure
• B. High temperature
• C. Decreasing $S{O}_3$ concentration
• D. Increasing reactant concentration

Answer 1

The correct option is B High temperature

The given reaction is
$2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$

We know, with increase in pressure the equilibrium will shift towards the side where fewer moles of the gases are present. So here at high pressure the rate of forward reaction will increase.

Since the reaction is exothermic, thus with increase in temperature the rate of backward reaction will increase. So high temperature is not favourable for the formation of $S{O}_3$.

Decreasing $S{O}_3$ concentration will shift the reaction towards forward direaction according to Le chatelier's principle. So it is favourable for $S{O}_3$ formation.

Increasing reactant concentration also increases the formation of $S{O}_3$ according to Le chatelier's principle only.